Question 7: Electrostats & Capacitance

Question

The electrostatic potential in a charged spherical region of radius $r$ varies as $V = ar^3 + b$, where $a$ and $b$ are constants. The total charge in the sphere of unit radius is $\alpha 	imes \pi a \epsilon_0$. The value of $\alpha$ is ______.(permittivity of vacuum is $\epsilon_0$)

Accepted Answer

The electric field $\vec{E}$ is related to the electric potential $V$ by:$\$$E = -\frac{dV}{dr}$$Since the potential $V$ depends only on the radial distance $r$, the electric field is radial. Given the potential function:$$V = ar^3 + b$\$Step 1: Find the Electric FieldThe electric field is calculated as:$$E = -\frac{dV}{dr} = -\left[ \frac{d}{dr}(ar^3 + b) ight] = -3ar^2$\$Step 2: Calculate Flux at Unit RadiusWe need to find the total charge enclosed in a sphere of unit radius ($R = 1$). The electric flux passing through the surface is:$\$\Phi_E = \oint \vec{E} \cdot d\vec{A}$$At $R = 1$, the electric field magnitude is:$$E(1) = -3a(1)^2 = -3a$$The surface area of the sphere is $4\pi R^2 = 4\pi(1)^2 = 4\pi$. Thus, the flux is:$$\Phi_E = E 	imes 	ext{Area} = (-3a) 	imes (4\pi) = -12\pi a$\$Step 3: Apply Gauss's LawAccording to Gauss's Law:$$\Phi_E = \frac{Q_{	ext{enclosed}}}{\epsilon_0} \implies Q_{	ext{enclosed}} = \Phi_E 	imes \epsilon_0$$$$\implies Q_{	ext{enclosed}} = (-12\pi a) 	imes \epsilon_0$$$$\implies Q_{	ext{enclosed}} = -12\pi a \epsilon_0$$By comparing this to the given form $Q = \alpha 	imes \pi a \epsilon_0$,$\$ we find:$$\alpha = -12$$

Answer

A. $-12$

Answer

B. $-8$

Answer

C. $-9$

Answer

D. $-6$

Electrostats & Capacitance

The electrostatic potential in a charged spherical region of radius r varies as V = ar^3 + b, where a and b are constants. The total charge in the sphere of unit radius is \alpha \times \pi a \epsilon_0. The value of \alpha is ______.(permittivity of vacuum is \epsilon_0)

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